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erica [24]
3 years ago
7

The air inside a hot-air balloon is heated to 59.00 °C (Tinitial) and then cools to 39.00 °C (Tfinal). By what percentage does t

he volume of the balloon change?
Chemistry
1 answer:
Mama L [17]3 years ago
4 0

Answer:

6.1%

Explanation:

Assuming pressure inside balloon remains constant during the temperature change.

Therefore, as per Charles' law  at constant pressure,

\frac{V_1}{T_1} =\frac{V_2}{T_2}

T_2=39\°\ C=39+273=312\ K

T_1=59°\ C=273+59=332\ K

V_2=V_1\times \frac{T_2}{T_1} \\=V_1\times\frac{312}{332}\\=0.939V_1

Percentage change in volume

\%\ change\ in\ volume=\frac{V_1-0.939V_1}{V_1} \times 100\\=6.1\%

Change in volume of the balloon is 6.1%

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How is percent by mass and concentration related?
Julli [10]
You can have a solution of hydrogen peroxide that might say 10% that means that 10% per mass of the hydrogen peroxide solution is the hydrogen peroxide the rest is water.

concentration is the amount of mass in the solution eg 5gdm-3

hope that helps
6 0
2 years ago
How many seconds would it take to deposit 17.3 g of ag (atomic mass = 107.87) from a solution of agno3 using a current of 10.00
Brums [2.3K]
Data Given:

Time = t = ?

Current = I = 10 A

Faradays Constant = F = 96500

Chemical equivalent = e = 107.86/1 = 107.86 g

Amount Deposited = W = 17.3 g

Solution:
             According to Faraday's Law,

                                          W  =  I t e / F

Solving for t,

                                          t  =  W F / I e
Putting values,
                                          t  =  (17.3 g × 96500) ÷ (10 A × 107.86 g)

                                          t  =  1547.79 s

                                          t  = 1.54 × 10³ s
5 0
2 years ago
A buffer contains 0.18 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. What is the pH
Yuki888 [10]

Answer:

1) pH = 5.05

2) pH = 5.13

3) pH = 4.97

Explanation:

Step 1: Data given

Number of moles of propionic acid = 0.18 moles

Number of moles sodium propionate = 0.26 moles

Volume = 1.20 L

Ka = 1.3 * 10^-5    → pKa = 4.989

Step 2: Calculate concentrations

Concentration = moles / volume

[acid]= 0.18/ 1.2 =0.150 M

[salt]= 0.26/ 1.3 = 0.217 M

pH = 4.89 + log(0.217/0.150)=<u>5.05</u>

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What is the pH of the buffer after the addition of 0.02 mol of NaOH?

moles acid = 0.18 - 0.02 = 0.16

[acid]= 0.16/ 1.2=0.133 M

moles salt = 0.26 + 0.02 = 0.28

[salt]= 0.28/ 12=0.233

pH = 4.89 + log 0.233/ 0.133 = 5.13

What is the pH of the buffer after the addition of 0.02 mol of HI?

moles acid = 0.18+ 0.02 = 0.20 moles

[acid]= 0.20/ 1.2 = 0.167 M

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8 0
2 years ago
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arlik [135]

Answer:

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Explanation:

8 0
3 years ago
Read 2 more answers
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olasank [31]

1) As can be seen from any 1H NMR chemical shift ppm tables, hydrogens which have δ values from 2ppm to 2.3ppm are hydrogens from carbon which is bonded to a carbonyl group. From this, we can conclude that our hydrogens belong to the type, but from 2 different alkyl groups because of 2 different signals.

 

2) So, one alkyl group is CH3 and second one can be CH or CH2.

 

3) If we know that ratio between two types of hydrogens is 3:2, it can be concluded that second alkyl group is CH2. 


4) Finally, we don't have any other signals and it indicates that part of the compound which continues on CH2 is exactly the same as the first part.

The ratio remains the same, 3:2 ie 6:4

7 0
3 years ago
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